Question

Solve the following systems of equations by the method of cross-multiplication:
$ax+by=a-b$
$bx-ay=a+b$

Answer 1

$ax+by=a-b$becomes

$\frac{(ax+by)}{(a-b)}=1$

$bx-ay=a+b$becomes

$\frac{(bx-ay)}{(a+b)}=1$

So now we can equate the two:

$\frac{(ax+by)}{(a-b)}=\frac{(bx-ay)}{(a+b)}$

So by cross multiplication:

$(ax+by)(a+b)=(bx-ay)(a-b)$

$a^{2}x+abx+aby+b^{2}y=abx-b^{2}x-a^{2}y+aby$

$a^{2}x+b^{2}y=-b^{2}x-a^{2}y$

$a^{2}x+b^{2}x=-a^{2}y-b^{2}y$

$(a^2+b^2)x=-(a^2+b^2)y$

$x=-y$

Substituting into the first equation:

$-ay+by=a-b$

$-(a-b)y=a-b$

$y=-1$

So $x=1$